Let $B subseteq biguplus^infty_n=1 A_n$, show that $mathbbP(B)=sum^infty_n=1 mathbbP(A_n) mathbbP(B|A_n)$Show: $mathbbE(f|mathcalF)=mathbbE(f)$Sequence of independent events in a discrete probability spaceLet $A_n$ be a sequence of Events. Find the Lim SupSigma additivity of events implies events being disjointSequence of events; infinitely often and sufficiently large $n$ exampleShow that the function $mathbbQ(A) = sum _ i:omega _ i in A ^,p_i$ is a probability measure on $(Omega, mathcalF)$Pairwise independence + additional condition imply independenceIf $mu$ is a vector measure and $(A_n)_n$ is disjoint, are we able to show that $(left|mu(A_n)right|)_n$ is summable$?Show that if $sum_n=1^inftymu(A_n) <infty$ we have that $mu(B) = 0$.Is $lim_ntoinfty mathbbP(A_n) = mathbbP(cup_n A_n)$?

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Let $B subseteq biguplus^infty_n=1 A_n$, show that $mathbbP(B)=sum^infty_n=1 mathbbP(A_n) mathbbP(B|A_n)$


Show: $mathbbE(f|mathcalF)=mathbbE(f)$Sequence of independent events in a discrete probability spaceLet $A_n$ be a sequence of Events. Find the Lim SupSigma additivity of events implies events being disjointSequence of events; infinitely often and sufficiently large $n$ exampleShow that the function $mathbbQ(A) = sum _ i:omega _ i in A ^,p_i$ is a probability measure on $(Omega, mathcalF)$Pairwise independence + additional condition imply independenceIf $mu$ is a vector measure and $(A_n)_n$ is disjoint, are we able to show that $(left|mu(A_n)right|)_n$ is summable$?Show that if $sum_n=1^inftymu(A_n) <infty$ we have that $mu(B) = 0$.Is $lim_ntoinfty mathbbP(A_n) = mathbbP(cup_n A_n)$?













1












$begingroup$



Question:



Let $(Omega, mathcalA, mathbbP)$ be a probability space with events
$A,BinmathcalA$. Now, let $B subseteq biguplus^infty_n=1 A_n$, where $A_n in mathcalA$ for each $ n in mathbbN$.



Show that: $mathbbP(B)=sum^infty_n=1 mathbbP(A_n) mathbbP(B|A_n)$




My attempt so far was to restructure the right-hand side to $sum^infty_n=1mathbbP(A_n)fracmathbbP(B cap A_n)mathbbP(A_n) = sum^infty_n=1mathbbP(B cap A_n)$ but I don't know if I'm even on the right track, so I'm pretty much stuck at this point.










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Mathician is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$











  • $begingroup$
    Caveat: you must suppose $P(A_n)ne 0$ for all $n$.
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Mar 10 at 20:18
















1












$begingroup$



Question:



Let $(Omega, mathcalA, mathbbP)$ be a probability space with events
$A,BinmathcalA$. Now, let $B subseteq biguplus^infty_n=1 A_n$, where $A_n in mathcalA$ for each $ n in mathbbN$.



Show that: $mathbbP(B)=sum^infty_n=1 mathbbP(A_n) mathbbP(B|A_n)$




My attempt so far was to restructure the right-hand side to $sum^infty_n=1mathbbP(A_n)fracmathbbP(B cap A_n)mathbbP(A_n) = sum^infty_n=1mathbbP(B cap A_n)$ but I don't know if I'm even on the right track, so I'm pretty much stuck at this point.










share|cite|improve this question







New contributor




Mathician is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    Caveat: you must suppose $P(A_n)ne 0$ for all $n$.
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Mar 10 at 20:18














1












1








1





$begingroup$



Question:



Let $(Omega, mathcalA, mathbbP)$ be a probability space with events
$A,BinmathcalA$. Now, let $B subseteq biguplus^infty_n=1 A_n$, where $A_n in mathcalA$ for each $ n in mathbbN$.



Show that: $mathbbP(B)=sum^infty_n=1 mathbbP(A_n) mathbbP(B|A_n)$




My attempt so far was to restructure the right-hand side to $sum^infty_n=1mathbbP(A_n)fracmathbbP(B cap A_n)mathbbP(A_n) = sum^infty_n=1mathbbP(B cap A_n)$ but I don't know if I'm even on the right track, so I'm pretty much stuck at this point.










share|cite|improve this question







New contributor




Mathician is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$





Question:



Let $(Omega, mathcalA, mathbbP)$ be a probability space with events
$A,BinmathcalA$. Now, let $B subseteq biguplus^infty_n=1 A_n$, where $A_n in mathcalA$ for each $ n in mathbbN$.



Show that: $mathbbP(B)=sum^infty_n=1 mathbbP(A_n) mathbbP(B|A_n)$




My attempt so far was to restructure the right-hand side to $sum^infty_n=1mathbbP(A_n)fracmathbbP(B cap A_n)mathbbP(A_n) = sum^infty_n=1mathbbP(B cap A_n)$ but I don't know if I'm even on the right track, so I'm pretty much stuck at this point.







probability-theory measure-theory conditional-probability






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Mathician is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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Mathician is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question




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asked Mar 10 at 19:52









MathicianMathician

104




104




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New contributor





Mathician is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Mathician is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











  • $begingroup$
    Caveat: you must suppose $P(A_n)ne 0$ for all $n$.
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Mar 10 at 20:18

















  • $begingroup$
    Caveat: you must suppose $P(A_n)ne 0$ for all $n$.
    $endgroup$
    – Martín-Blas Pérez Pinilla
    Mar 10 at 20:18
















$begingroup$
Caveat: you must suppose $P(A_n)ne 0$ for all $n$.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 10 at 20:18





$begingroup$
Caveat: you must suppose $P(A_n)ne 0$ for all $n$.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 10 at 20:18











1 Answer
1






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oldest

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1












$begingroup$

Hint: if $biguplus$ means disjoint union,
$$B = Bcapleft(biguplus_n=1^infty A_nright) =
biguplus_n=1^infty(Bcap A_n).
$$






share|cite|improve this answer









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    1












    $begingroup$

    Hint: if $biguplus$ means disjoint union,
    $$B = Bcapleft(biguplus_n=1^infty A_nright) =
    biguplus_n=1^infty(Bcap A_n).
    $$






    share|cite|improve this answer









    $endgroup$

















      1












      $begingroup$

      Hint: if $biguplus$ means disjoint union,
      $$B = Bcapleft(biguplus_n=1^infty A_nright) =
      biguplus_n=1^infty(Bcap A_n).
      $$






      share|cite|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$

        Hint: if $biguplus$ means disjoint union,
        $$B = Bcapleft(biguplus_n=1^infty A_nright) =
        biguplus_n=1^infty(Bcap A_n).
        $$






        share|cite|improve this answer









        $endgroup$



        Hint: if $biguplus$ means disjoint union,
        $$B = Bcapleft(biguplus_n=1^infty A_nright) =
        biguplus_n=1^infty(Bcap A_n).
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 10 at 20:15









        Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

        34.6k42971




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